Chinese Remainder Theorem (Commutative Algebra)/Proof 2
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Theorem
Let $A$ be a commutative and unitary ring.
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Let $I_1, \ldots, I_n$ for some $n \ge 1$ be ideals of $A$.
Then the ring homomorphism $\phi: A \to A / I_1 \times \cdots \times A / I_n$ defined as:
- $\map \phi x = \tuple {x + I_1, \ldots, x + I_n}$
has the kernel $\ds I := \bigcap_{i \mathop = 1}^n I_i$, and is surjective if and only if the ideals are pairwise coprime, that is:
- $\forall i \ne j: I_i + I_j = A$
Hence in that case, it induces an ring isomorphism:
- $A / I \to A / I_1 \times \cdots \times A / I_n$
through the First Isomorphism Theorem.
Proof
Consider $\pi$ only as a homomorphism of groups.
Then Chinese Remainder Theorem (Groups) is applicable as Subgroup of Abelian Group is Normal.
It remains to demonstrate that the condition $I_i + I_j = R$ for all $i \ne j$ assumed here is equivalent to:
- $\ds \forall k \le n - 1: I_{k + 1} + \bigcap_{i \mathop = 1}^k I_i = R$
The implication from the latter condition is immediate.
For the converse, let $i$ be arbitrary.
The result follows from:
\(\ds R\) | \(=\) | \(\ds \prod_{j \mathop \ne i} \paren {I_i + I_j}\) | $R$ has a unit and $I_i + I_j = R$ for $j \ne i$ | |||||||||||
\(\ds \) | \(\subseteq\) | \(\ds I_i + \prod_{j \mathop \ne i} I_j\) | Ring Axiom $\text D$: Distributivity of Product over Addition, Definition of Ideal of Ring | |||||||||||
\(\ds \) | \(\subseteq\) | \(\ds I_i + \bigcap_{j \mathop \ne i}^n I_j\) | Intersection of Ideals of Ring contains Product |
$\blacksquare$
Sources
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- 1990: Kenneth Ireland and Michael Rosen: A Classical Introduction to Modern Number Theory (2nd ed.): Chapter $12$ Algebraic Number Theory: $\S 3$ Ramification and Degree: Proposition $12.3.1$
- 2005: Serge Lang: Undergraduate Algebra (3rd ed.): Chapter $\text {III}$, $\S 3$ Exercises: Problem $7$
- 2012: Ambar N. Sengupta: Representing Finite Groups: A Semisimple Introduction: Proposition $12.4.1$